would anyone be able to give me a hint to get me going with this problem, i am unsure as what i should start by doing... small hint?

 

Cheers

 

Sarah

i have figured out how to solve it exactly (its 3/2), but 10^-6 stuff, i dunno :S

ok, this is where i have started:

You want an N so that:

 

[math]\sum_{n=1}^{\infty}\frac{n^2}{3^n}-P_N<10^{-6}[/math]

 

where [math]P_N=\sum_{n=1}^{N}\frac{n^2}{3^n}[/math] (there's no dependance on "x" here). The left side is just a sum from N+1 to infinity. Use their hint and dominate this sum with a geometric series to get an upper bound dependant on N. Choose N appropriately.

hang on, my computer it just showing little boxes with ? for the latex stuff, damn my computer! i'll go to another one

hmm ok so far it makes sense sort of, i have tryed manually trying to find a value for N, but obviously that is not going to work, but i can see that its heading in the right direction...

ok so i will bound it with the series

 

1+2+4+8+16+....

 

which is geometric series with a=1 and a comon ratio r=2

ok now i've got this, (after using the formula for the partial sum of a geometric series applied to the series i mentioned in my previous post)

well i'm stuck dammit

ok so, so far this is whats happenin:

[math]

\sum_{n=1}^{\infty}\frac{n^2}{3^n}-\sum_{n=1}^{N}\frac{n^2}{3^n}<10^{-6}

=\sum_{n=N+1}^{\infty}\frac{n^2}{3^n}

[/math]

ok this is the way i think to go is

[math]

\sum_{n=1}^{N}\frac{n^2}{3^n}>\sum_{n=1}^{\infty}\frac{n^2}{3^n}-10^{-6}

[/math]

then you somehow find out what that N value is, because you know that

[math]

\sum_{n=1}^{\infty}\frac{n^2}{3^n}=\frac{3}{2}

[/math]

mmm?

 

you want an N such that

 

[math] \sum_{r=N+1}^{\infty} r^2/3^r[/math]

 

is less than 10^{-6}

 

Well, that sum is less than

 

[math]\sum_{r=N+1}^{\infty}(2/3)^r[/math]

 

 

for N sufficiently large (ie when 2^r>r^2 which is true for all r>5 certainly)

 

 

and that sum equals 3(2/3)^{N+1}

 

so pick N such that that is less than than 10^{-6}

 

say any N greater than 100 would do

hmm ok, i thought of another method, which sort of makes sense.

we need to find a N s.t

[math]

\sum_{r=N+1}^{\infty} r^2/3^r<10^{-6}

[/math]

since each term is <= the last one in this series, then if you can make the first term <10^{-6} then you can easily come up with a value of N. (i got 17 or 18).

 

hmm but then i suppose you could have a series which goes 1, 0.8, 0.5, 0.1,0.1,0.1... and so yeah if you kept adding those all up i suppose yeah, ok i think what i just said above is not correct

yes i see now, well it works for r>=5, but thats not important (my only ever chance to correct you! )

 

hmm ok now, we want:

[math]

3(\frac{2}{3})^{N+1}<10^{-6}

[/math]

 

and so any value of N>=36 will do

 

yep i think i understand this now, i knew what it was asking, i just didnt quite know how to implement their hint,

 

 

hehe i think i've got the hang of this latex thing now

 

thanks matt

hmm ok' date=' i thought of another method, which sort of makes sense.

we need to find a N s.t

[math']

\sum_{r=N+1}^{\infty} r^2/3^r<10^{-6}

[/math]

since each term is <= the last one in this series, then if you can make the first term <10^{-6} then you can easily come up with a value of N. (i got 17 or 18).

... i suppose yeah, ok i think what i just said above is not correct

 

 

certainly it is not correct.

yes i see now, well it works for r>=5, but thats not important

 

sorry, but you aren't correcting me. i didnt' say that r>5 was neccesary, only sufficient.

sorry, but you aren't correcting me. i didnt' say that r>5 was neccesary, only sufficient.

 

lol, my dreams crushed! haha i was a fool to think i would ever be able to correct you

but it is an important point: do you understand the difference between "if", "only if", and "if and only if"? sometimes they use sufficient, necessary, and necessary and sufficient.

oh ok, yep, i see.

hmmm...in the hint it says to use the comparison theorem. i don't see why we would really need to, but anyway

because the series you can't sum r^2/3^n you replace by one you can, (2/3)^n and you compare the (easily evaluated) sum of the tail of one and which by the comparison theorem we know is more than the sum of the tail of the other.

oh right, yep of course, sorry yeah i get it now

 

thanks matt